-0.000 000 000 847 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 847(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 847(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 847| = 0.000 000 000 847


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 847.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 847 × 2 = 0 + 0.000 000 001 694;
  • 2) 0.000 000 001 694 × 2 = 0 + 0.000 000 003 388;
  • 3) 0.000 000 003 388 × 2 = 0 + 0.000 000 006 776;
  • 4) 0.000 000 006 776 × 2 = 0 + 0.000 000 013 552;
  • 5) 0.000 000 013 552 × 2 = 0 + 0.000 000 027 104;
  • 6) 0.000 000 027 104 × 2 = 0 + 0.000 000 054 208;
  • 7) 0.000 000 054 208 × 2 = 0 + 0.000 000 108 416;
  • 8) 0.000 000 108 416 × 2 = 0 + 0.000 000 216 832;
  • 9) 0.000 000 216 832 × 2 = 0 + 0.000 000 433 664;
  • 10) 0.000 000 433 664 × 2 = 0 + 0.000 000 867 328;
  • 11) 0.000 000 867 328 × 2 = 0 + 0.000 001 734 656;
  • 12) 0.000 001 734 656 × 2 = 0 + 0.000 003 469 312;
  • 13) 0.000 003 469 312 × 2 = 0 + 0.000 006 938 624;
  • 14) 0.000 006 938 624 × 2 = 0 + 0.000 013 877 248;
  • 15) 0.000 013 877 248 × 2 = 0 + 0.000 027 754 496;
  • 16) 0.000 027 754 496 × 2 = 0 + 0.000 055 508 992;
  • 17) 0.000 055 508 992 × 2 = 0 + 0.000 111 017 984;
  • 18) 0.000 111 017 984 × 2 = 0 + 0.000 222 035 968;
  • 19) 0.000 222 035 968 × 2 = 0 + 0.000 444 071 936;
  • 20) 0.000 444 071 936 × 2 = 0 + 0.000 888 143 872;
  • 21) 0.000 888 143 872 × 2 = 0 + 0.001 776 287 744;
  • 22) 0.001 776 287 744 × 2 = 0 + 0.003 552 575 488;
  • 23) 0.003 552 575 488 × 2 = 0 + 0.007 105 150 976;
  • 24) 0.007 105 150 976 × 2 = 0 + 0.014 210 301 952;
  • 25) 0.014 210 301 952 × 2 = 0 + 0.028 420 603 904;
  • 26) 0.028 420 603 904 × 2 = 0 + 0.056 841 207 808;
  • 27) 0.056 841 207 808 × 2 = 0 + 0.113 682 415 616;
  • 28) 0.113 682 415 616 × 2 = 0 + 0.227 364 831 232;
  • 29) 0.227 364 831 232 × 2 = 0 + 0.454 729 662 464;
  • 30) 0.454 729 662 464 × 2 = 0 + 0.909 459 324 928;
  • 31) 0.909 459 324 928 × 2 = 1 + 0.818 918 649 856;
  • 32) 0.818 918 649 856 × 2 = 1 + 0.637 837 299 712;
  • 33) 0.637 837 299 712 × 2 = 1 + 0.275 674 599 424;
  • 34) 0.275 674 599 424 × 2 = 0 + 0.551 349 198 848;
  • 35) 0.551 349 198 848 × 2 = 1 + 0.102 698 397 696;
  • 36) 0.102 698 397 696 × 2 = 0 + 0.205 396 795 392;
  • 37) 0.205 396 795 392 × 2 = 0 + 0.410 793 590 784;
  • 38) 0.410 793 590 784 × 2 = 0 + 0.821 587 181 568;
  • 39) 0.821 587 181 568 × 2 = 1 + 0.643 174 363 136;
  • 40) 0.643 174 363 136 × 2 = 1 + 0.286 348 726 272;
  • 41) 0.286 348 726 272 × 2 = 0 + 0.572 697 452 544;
  • 42) 0.572 697 452 544 × 2 = 1 + 0.145 394 905 088;
  • 43) 0.145 394 905 088 × 2 = 0 + 0.290 789 810 176;
  • 44) 0.290 789 810 176 × 2 = 0 + 0.581 579 620 352;
  • 45) 0.581 579 620 352 × 2 = 1 + 0.163 159 240 704;
  • 46) 0.163 159 240 704 × 2 = 0 + 0.326 318 481 408;
  • 47) 0.326 318 481 408 × 2 = 0 + 0.652 636 962 816;
  • 48) 0.652 636 962 816 × 2 = 1 + 0.305 273 925 632;
  • 49) 0.305 273 925 632 × 2 = 0 + 0.610 547 851 264;
  • 50) 0.610 547 851 264 × 2 = 1 + 0.221 095 702 528;
  • 51) 0.221 095 702 528 × 2 = 0 + 0.442 191 405 056;
  • 52) 0.442 191 405 056 × 2 = 0 + 0.884 382 810 112;
  • 53) 0.884 382 810 112 × 2 = 1 + 0.768 765 620 224;
  • 54) 0.768 765 620 224 × 2 = 1 + 0.537 531 240 448;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 847(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1010 0011 0100 1001 0100 11(2)

6. Positive number before normalization:

0.000 000 000 847(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1010 0011 0100 1001 0100 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 847(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1010 0011 0100 1001 0100 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1010 0011 0100 1001 0100 11(2) × 20 =

1.1101 0001 1010 0100 1010 011(2) × 2-31


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1101 0001 1010 0100 1010 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 110 1000 1101 0010 0101 0011 =

110 1000 1101 0010 0101 0011


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
110 1000 1101 0010 0101 0011


Decimal number -0.000 000 000 847 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 110 1000 1101 0010 0101 0011

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111